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Numerade Educator



Problem 19 Easy Difficulty

Find the limit or show that it does not exist.

$ \displaystyle \lim_{t \to \infty}\frac{\sqrt{t} + t^2}{2t - t^2} $



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Video Transcript

All right. So we want to find this limit or show that does not exist. Um And as usual, some people like to do low petals rule here or talk about end behavior. Um I think the best way uh to do limits is to use algebra because limits have to exist before um before derivatives do so lumpy tiles rule doesn't make any sense. You may or may not know that already. So this is really big. The top is really big. The bottom is really big or maybe negative. Really big. I can't tell I'm just going to stop things from being big. So I'm going to divide top and bottom by T squared because that's the thing that's getting big the fastest. Okay, so what happens when I do that? Well, T T is like T to the one half. Right. So this is Um won over now. T to the 3/2. If you just use expanded rules. Um you can think of it as teeth. The negative three halves and then write it as 1/2 to the three halves. Okay, Plus T squared over T squared is one divided by um to T over T squared is two over tea and then minus T squared over two squared is one. I'm just going to write this one more time as something slightly different. This three halves is kind of strange. So one over T to the three halves plus one. And why do I want this? Usually don't want fractions infractions, but um, I only need one theorem for something getting big as T gets big one over T gets small, right one divided by a huge number is always going to be really, really small, as the number gets huger and huger, that goes towards zero, that's what that means. Okay, now, I just have that theorem everywhere in here, I have one divided by big number small, raised, the power of three halves is still small, so that limit, the limit of that part is zero, so then the top is zero plus one is one, divided by the bottom is two times this part goes to zero minus one, it's negative. So my answer is negative one, and I think that's the cleanest way to think about a limit like this.